OFFSET
0
COMMENTS
Let psi_8 be the elementary Sturmian morphism given by
psi_8(0)=01, psi(8)1=1,
and let x = A003849 be the Fibonacci word. Then, by definition, (a(n)) = psi_8(x). Now x is a Sturmian sequence x = s(alpha, rho) with slope alpha = (3-sqrt(5))/2 and intercept rho = alpha. This implies that (a(n)) is a Sturmian sequence with slope
alpha' = 1/(2-alpha) = (sqrt(5)-1)/2,
and intercept
rho' = rho/(2-alpha) = sqrt(5)-2 (cf. Lothaire Lemma 2.2.18).
Since the algebraic conjugate of rho' is equal to -sqrt(5)-2, which is smaller than the algebraic conjugate of alpha', (a(n)) is NOT a fixed point of a morphism, by Yasutomi's criterion. However, (a(n)) IS a fixed point of an automorphism sigma of the free group generated by 0 and 1.
To see this, let gamma be the Fibonacci morphism given by
gamma(0)=01, gamma(1)=0.
Then gamma(x) = x, and so
psi_8(gamma(x)) = psi_8(x) = a,
implying that a = (a(n)) is fixed by
sigma:=psi_8 gamma psi_8^{-1}.
One easily computes psi_8^{-1}: 0->01^{-1}, 1->1, which gives sigma:
sigma(0) = 010^{-1}, sigma(1) = 01.
The binary complement of (a(n)) is equal to the left shift of the Fibonacci sequence, i.e., if c(n):= 1-a(n), then c(n)=A003849(n+1), for n=0,1,2,....
This can be proved by observing that (A003849(n+1)) is again a Sturmian sequence with slope alpha and intercept 2*alpha, and (c(n)) is again a Sturmian sequence with slope 1-alpha' = (3-sqrt(5)/2 and intercept 1- rho' = 3-sqrt(5) (see Lothaire Lemma 2.2.17).
LINKS
M. Lothaire, Algebraic combinatorics on words, Cambridge University Press. Online publication date: April 2013; Print publication year: 2002.
FORMULA
a(n) = ((n+1)*alpha' + rho') - (n*alpha' + rho'),
where alpha' = (sqrt(5)-1)/2, and rho' = sqrt(5)-2.
EXAMPLE
To see that (a(n)) is fixed by sigma, iterate sigma starting with 01.
sigma(01) = 010^{-1}01 = 011,
sigma^2(01) = 010^{-1}0101 = 01101,
sigma^3(01) = 010^{-1}0101010^{-1}01 = 01101011.
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Dekking, May 29 2018
STATUS
approved